Alex Blum wrote:
>Why does it follow that if there is no model of ZFC
>in which con(ZFC)
> is
>true that the negation of con(ZFC) has a proof in
>ZFC?
>Doesn't this assume that ZFC is complete?
No, it doesn't.
For any first order theory T, T has a model iff T is
consistent (here I had previously mentioned Henkin's
1949 result but Martin Davis has reminded me of the
fact that this was already in Gödel's 1930
completeness theorem); for any theory T and
any sentence S, T+S is inconsistent iff the negation
of S is a logical consequence of T.
Now, since first order logic is complete, ZFC proves
whatever is a logical consequence of its axioms. So,
if ZFC+Con(ZFC) has no model, then ZFC proves the
negation of Con(ZFC).
Laureano Luna
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